The given problem asks for the derivative of the function $y = 5 \cos(3x)$. Let's solve this step-by-step:

Step-by-Step Solution:

1. Given Function:
   $y = 5 \cos(3x)$

2. Apply the Derivative Rule:
   The derivative of $\cos(u)$ is $-\sin(u)$, where $u$ is a function of $x$. Using the chain rule: [ y' = 5 \cdot \frac{d}{dx}[\cos(3x)] = 5 \cdot (-\sin(3x)) \cdot \frac{d}{dx}[3x] ]

3. Differentiate the Inner Function:
   $\frac{d}{dx}[3x] = 3$

4. Combine Results:
   Substituting back: $y' = 5 \cdot (-\sin(3x)) \cdot 3 = -15 \sin(3x)$

Final Answer:

The derivative is: $y' = -15 \sin(3x)$

Thus, the correct option is: $\text{d) } y' = -15 \sin(3x)$

Would you like a more detailed explanation of the chain rule or trigonometric derivatives?

Here are 5 related questions to expand your understanding:

1. What is the derivative of $y = \sin(5x)$?
2. How does the chain rule apply to composite functions like $\cos(3x)$?
3. Can you explain the difference between the derivative of $\sin(x)$ and $\cos(x)$?
4. What is the integral of $y = -15 \sin(3x)$?
5. How do we handle higher-order derivatives of trigonometric functions?

Tip: Always verify your differentiation by checking intermediate steps, especially when applying the chain rule!